Deriving formula by solving the general form of quadratic equation 'Ax2 + Bx + C = 0' by the method of completing the square ..

STEP-[1]- Substract the constant term from both sides of the quadratic equation, or just remove the constant term from the left hand side of equation ,change the sign of it
,if constant term is positive change it to negative if it is negative change it to positive and put it on the right hand side ,removing the original 0 from the right hand side .i.e Ax2 + Bx + C - C = 0 - C
Ax2 + Bx + 0 = -C
Ax2 + Bx = -C

STEP-[2]- divide throughout by the numerical coefficient of the quadratic term or the coefficient of x2 ;
i.e , divide entire equation by 'A' the number in front of x2 .
now since 1x2 = 1 times x2 = x2, when you see no number in front of
x2 know that 1 is really the number in front of x2 or that 1 is the
coefficient of x2,
further since 4/1 = 4 , y/1 = y etc.. more generally ,'any number divided by 1 gives the result of the number itself'. when you see no number in front of x2 ,you now know that 1 is the coefficient of x2 donot bother to divide by 1 ,just skip this step.
Ax2 + Bx = -C
( Ax2½A ) + Bx½A = -C ½A
x2 + (B½A)x = -C ½A

STEP-[3]-Now find half of the coefficient of x ,i.e find half of the number now in front of x ,to do that just multiply 1/2 to the coefficient of x ,since the coefficient of x is B/A ,we have ; (1/2)*B/A = B/2A

STEP-[4]-Now square half of the coefficient of x ,i.e square the result obtained in step[3] i.e is (B/2A)2

STEP-[5]-Now add the square of half of the coefficient of x ,i.e
the result obtained in step[4] to both sides of equation obtained in step[2] i.e
x2 + (B½A)x + (B/2A)2 = (-C ½A)+ (B/2A)2
Now factorising the left hand side ,
since (B/2A)*(B/2A) = (B/2A)2
and (B/2A)+(B/2A) = B/A
replacing (B/A)x with (B/2A)x+(B/2A)x in the equation immediately above
we have
x2 + (B/2A)x +(B/2A)x + (B/2A)2 = (-C ½A)+ (B/2A)2
putting the terms on the left hand side into groups of two
we have
[ x2 + (B/2A)x ] + [(B/2A)x + (B/2A)2 ] = (-C ½A)+ (B/2A)2
now factor each group on left hand side
x[x + (B/2A)] + (B/2A)[ x + (B/2A)] = (-C ½A)+ (B/2A)2
now factorise out x + (B/2A) from left hand side
[x + (B/2A)][x + (B/2A)] = (-C ½A)+ (B/2A)2
which gives :
[x + (B/2A)]2 =(-C ½A)+ (B/2A)2
Note : since x2 + (B½A)x + (B/2A)2 is a perfect square x2 + (B½A)x + (B/2A)2 = [x + (B/2A)] 2
you donot have to factorise the Left hand side as I did ,by factorising I actually
proved why x2 + (B½A)x + (B/2A)2 = [x + (B/2A)] 2
when you are using the method of completing the square skip the factorising bit.
Now to solve for x take the square root of both sides of equation
since the square root of [x + (B/2A)] 2 is [x + (B/2A)]
we have ;
x + (B/2A) = (+/-) Ö [(-C ½A)+ (B/2A)2]
subtracting B/2A from both sides of equation immediately above solving for x
we have ; x = (-B/2A)(+/-)Ö [(-C ½A)+ (B/2A)2]
simplifying the right hand side of equation gives x = [ -B +/- Ö (B2 - 4AC) ] /2A